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2 years ago

ANSWER ASAP PLZ

Mathematics

2 answers:

Helen [10]2 years ago

5 0

Choice B is the answer because:
2*2=4
4*2=8
8*2=16
16*2=32
32*2=64
or
2^6= 64

chubhunter [2.5K]2 years ago

4 0

The answer is B... You can multiply 2*2*2*2*2*2, or you can do $2^{6}$ which ever way is easier to you.

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Can someone explain this and help me write it and write the list of steps

Naddik [55]

113 = 72 + m < B ( by the external angle of a triangle theorem)

Subtract 72 from both sides:-

113 - 72 = 72 - 72 + m < b

m<b = 41 degrees

6 0

3 years ago

How many solutions are in 17x-8=3x+16

Marat540 [252]

There is only one solution that satisfies the equation.
14x = 24
x = 24/14 = 12/7

4 0

1 year ago

Read 2 more answers

A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1491 and the standard deviati

Alinara [238K]

Answer:

$z(1910)=1.3429\\\\z(1230)=-0.9006\\\\z(2190)=2.2404\\\\z(1380)=-0.3558$

Step-by-step explanation:

-Given the population mean is $\mu=1491$ and the standard deviation is $\sigma=312$ , we calculate the z-scores using the formula:

$z=\frac{\bar x-\mu}{\sigma}$

#The z-score for 1910 can be calculated as:

$z(1910)=\frac{x-\mu}{\sigma}\\\\=\frac{1910-1491}{312}\\\\=\frac{419}{312}\\\\=1.3429$

#The z-score for 1230can be calculated as:

$z(1230)=\frac{x-\mu}{\sigma}\\\\=\frac{1230-1491}{312}\\\\=\frac{-281}{312}\\\\=-0.9006$

#The z-score for 2190 is calculated as follows:

$z(2190)=\frac{x-\mu}{\sigma}\\\\=\frac{2190-1491}{312}\\\\=\frac{699}{312}\\\\=2.2404$

#The z-score for 1380 is calculated as:

$z(1380)=\frac{x-\mu}{\sigma}\\\\=\frac{1380-1491}{312}\\\\=\frac{-111}{312}\\\\=-0.3558$

7 0

2 years ago

35 (t + 29) = −3(3 − t)

Olin [163]

Answer: -32

Step-by-step explanation:

35 (t + 29) = −3(3 − t)

Open the brackets

35t+1015= -9+3t

Collect like terms

35t-3t=-9-1015

32t= -1024

Divide both side by 32

32t/32 = -1024/32

t= -32

8 0

3 years ago

Read 2 more answers

Michele correctly solved a quadratic equation using the quadratic formula as shown below.

Fofino [41]

Answer:

The original equation must have been:

$7x^{2} -5x-2=0$

Step-by-step explanation:

A quadratic equation is in the form of :

$ax^{2} +bx+c=0$

And it is solved in the form of :

$x=\frac{-b+\sqrt{b^{-4ac} } }{2a}$ and $x=\frac{-b-\sqrt{b^{-4ac} } }{2a}$

Now the given equation is :

$x=\frac{-(-5)+-\sqrt{(-5)^{2} -4(7)(-2)} }{2(7)}$

We can see that here;

a = 7

b = -5

c = -2

So, the original equation must have been:

$7x^{2} -5x-2=0$

5 0

3 years ago